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Chapter 5: Problem 4

Multiply each expression using the product rule. $$y \cdot y^{19}$$

Short Answer

Expert verified
The simplified expression is \(y^{20}\)

Step by step solution

01

Express the first term with an exponent

First, write the expression \(y\) as \(y^{1}\). This means our expression now reads as \(y^{1} \cdot y^{19}\).
02

Apply the product rule

Apply the product rule of exponents. This means we add the exponents of both terms.
03

Simplify the expression

Finally, simplifying the expression, we get \(y^{1 + 19} = y^{20}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule of Exponents
In mathematics, the Product Rule of Exponents is an essential concept that simplifies working with exponents when multiplying similar bases. Let's break it down to make it more understandable.

When you have two expressions with the same base, for example, \(a^m \cdot a^n\), the product rule tells us that we can add the exponents, resulting in \(a^{m+n}\).
  • For instance, \(x^2 \cdot x^3\) becomes \(x^{2+3}\) which simplifies to \(x^5\).
  • This works because of the definition of exponents: multiplying \(x^2\) means \(x \times x\), and multiplying that by \(x^3\) (which is \(x \times x \times x\)) results in a total of five \(x's\) being multiplied together.
This rule significantly reduces the complexity of multiplying exponential terms by transforming multiplication into a simpler operation鈥攁ddition鈥攐nce the bases are equal.
Simplifying Expressions
Simplifying expressions is a vital step in solving algebraic problems and making equations easier to read and understand. The process involves reducing complicated expressions to their simplest form.

The main goal here is to make the expressions easier to work with, particularly when dealing with complex equations or further algebraic operations. Let's use our example:
  • First, we consider the expression \(y \cdot y^{19}\).
  • By recognizing \(y\) as \(y^1\), we rewrite it as \(y^1 \cdot y^{19}\).
  • Applying the product rule, we simplify this to \(y^{1+19}\).
  • The final simplification yields \(y^{20}\).
Each step is about reducing complexity by combining like terms using the fundamental rules of arithmetic and algebra. By mastering these simplification techniques, you can tackle more challenging problems with ease.
Algebra Basics
Algebra is the language of mathematics, which gives us the tools to express and solve relationships using symbols and letters to represent numbers and quantities.

Some essential basics of algebra relevant to our task include:
  • Variables: These are symbols (like \(x\), \(y\), or \(z\)) representing numbers whose exact values are not yet known. For example, in \(y^1 \cdot y^{19}\), \(y\) is a variable.
  • Exponents: These tell how many times to use the number (or variable) in a multiplication. For example, \(y^{19}\) means multiplying \(y\) by itself 19 times.
  • Expression: A combination of numbers, variables, and operators (like \(+\), \(-\), \(\times\), \(\div\)) that represents a particular value or relationship. \(y^1 \cdot y^{19}\) is an example of an algebraic expression.
Understanding these basics is crucial because they form the foundation for grasping more complex topics in mathematics. Armed with this knowledge, you can effectively approach and solve algebraic problems, making mathematics a more manageable and less intimidating subject.

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Most popular questions from this chapter

Explain how to divide a polynomial that is not a monomial by a monomial. Give an example.

Subtract: \(-4.6-(-10.2)\)

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\left(6 x^{2} y-7 x y-4\right)-\left(6 x^{2} y+7 x y-4\right)=0$$

Explain the difference between \((-7)^{0}\) and \(-7^{0}\).

Exercises \(172-174\) will help you prepare for the material covered in the first section of the next chapter. In each exercise, find the product. $$9 x y\left(3 x y^{2}-y+9\right)$$
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